Re: [sage-devel] Unimodular transformation matrix of LLL algorithm

2020-09-27 Thread Nils Bruin 


On Sunday, September 27, 2020 at 1:48:03 PM UTC-7, Santanu Sarkar wrote:
>
> Dear Nils,
>Thank you so much for your comments.  
> I consider Matrix E=[I,M1], where I is identity matrix. 
> Then reduction of E took 100 seconds. Hence I am not 
> going any advantage. 
>
> Try [10^b*M1,I] with b = 50 or so (it depends on your problem what a 
"large" number is). I'm not promising that it's faster, but it's worth a 
try. If you do not scale M1, then I would expect you're getting back 
different answers, unless all the vectors in the span of M1 already have 
very large norm.

I now also see that you're looking at a very "overgenerated" lattice (span 
of 90 vectors in ZZ^6?) Such a lattice almost surely has a very small 
covolume. I'd expect that just a HNF is actually quite fast to compute. The 
transformation matrix in such a case is also highly non-unique. It's not so 
surprising that keeping track of it is very much more expensive in this 
case (although I'm not sure it explains the full factor that you see).

I'd expect that something like 8 combinations of the 90 generating vectors 
already generate the same lattice. Now you're looking at LLL-reducing a 8x6 
matrix rather than a 90x6 matrix!



 

> Regards,
> Santanu 
>
>
> On Mon, 28 Sep 2020 at 01:12, Nils Bruin > 
> wrote:
>
>> You could do the same thing as you do with Gaussian elimination to track 
>> the row operations: augment the matrix with an identity matrix.
>> In order for the augmentation to not affect your LLL reduction, you'd 
>> want to multiply your original matrix by a large constant, so that the 
>> augmented coordinates do not affect the norms significantly.
>>
>> On Sunday, September 27, 2020 at 11:20:28 AM UTC-7, Dima Pasechnik wrote:
>>>
>>>
>>>
>>> On Sun, 27 Sep 2020, 18:43 Santanu Sarkar,  
>>> wrote:
>>>
 Dear Martin,
 Thank you so much. It works!
 Can we make it faster? 
 It took 17 seconds for my problem but 
 M1.LLL() took only 3 seconds. Of course I understand we 
 are calculating extra matrix U.  

>>> one needs to do some Cython programming, as suggested by
>>> the trac ticket mentioned above.
>>>
>>>
>>>
 Thanks again for your help. 

 Regards,
 Santanu 



 On Sun, 27 Sep 2020 at 20:45, 'Martin R. Albrecht' via sage-devel <
 sage-...@googlegroups.com> wrote:

> Hi there,
>
> This should do the trick:
>
> sage: from fpylll import *
> sage: A = random_matrix(ZZ, 6, 90)
> sage: U = IntegerMatrix.identity(6)
> sage: B = LLL.reduction(IntegerMatrix.from_matrix(A), 
> U).to_matrix(matrix(ZZ, 6,
> 90))
> sage: U = U.to_matrix(matrix(ZZ, 6,6))
> sage: B == U*A
> True
> sage: abs(U.det())
> 1
>
> Cheers,
> Martin
>
>
> Santanu Sarkar  writes:
> > Dear all,
> >I have a matrix M1 with integer entries with 90 rows and 6 
> columns.
> > After applying LLL algorithm of M1, I get M2=M1.LLL(). I want to get
> > corresponding unimodular transformation matrix T such that
> > T*M1=M2. We can find T by
> > T=M2*M1.pseudoinverse() or T== M1.solve_left(M2), but determinant of 
> T
> > becomes 0 i.e.,  T.det()=0.
> > I want T.det()=1.
> >
> > Best regards,
> > Santanu
>
>
> -- 
>
> _pgp: https://keybase.io/martinralbrecht
> _www: https://malb.io
>
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> .
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Re: [sage-devel] Unimodular transformation matrix of LLL algorithm

2020-09-27 Thread Santanu Sarkar 
Dear Nils,
   Thank you so much for your comments.
I consider Matrix E=[I,M1], where I is identity matrix.
Then reduction of E took 100 seconds. Hence I am not
going any advantage.

Regards,
Santanu


On Mon, 28 Sep 2020 at 01:12, Nils Bruin  wrote:

> You could do the same thing as you do with Gaussian elimination to track
> the row operations: augment the matrix with an identity matrix.
> In order for the augmentation to not affect your LLL reduction, you'd want
> to multiply your original matrix by a large constant, so that the augmented
> coordinates do not affect the norms significantly.
>
> On Sunday, September 27, 2020 at 11:20:28 AM UTC-7, Dima Pasechnik wrote:
>>
>>
>>
>> On Sun, 27 Sep 2020, 18:43 Santanu Sarkar,  wrote:
>>
>>> Dear Martin,
>>> Thank you so much. It works!
>>> Can we make it faster?
>>> It took 17 seconds for my problem but
>>> M1.LLL() took only 3 seconds. Of course I understand we
>>> are calculating extra matrix U.
>>>
>> one needs to do some Cython programming, as suggested by
>> the trac ticket mentioned above.
>>
>>
>>
>>> Thanks again for your help.
>>>
>>> Regards,
>>> Santanu
>>>
>>>
>>>
>>> On Sun, 27 Sep 2020 at 20:45, 'Martin R. Albrecht' via sage-devel <
>>> sage-...@googlegroups.com> wrote:
>>>
 Hi there,

 This should do the trick:

 sage: from fpylll import *
 sage: A = random_matrix(ZZ, 6, 90)
 sage: U = IntegerMatrix.identity(6)
 sage: B = LLL.reduction(IntegerMatrix.from_matrix(A),
 U).to_matrix(matrix(ZZ, 6,
 90))
 sage: U = U.to_matrix(matrix(ZZ, 6,6))
 sage: B == U*A
 True
 sage: abs(U.det())
 1

 Cheers,
 Martin


 Santanu Sarkar  writes:
 > Dear all,
 >I have a matrix M1 with integer entries with 90 rows and 6 columns.
 > After applying LLL algorithm of M1, I get M2=M1.LLL(). I want to get
 > corresponding unimodular transformation matrix T such that
 > T*M1=M2. We can find T by
 > T=M2*M1.pseudoinverse() or T== M1.solve_left(M2), but determinant of T
 > becomes 0 i.e.,  T.det()=0.
 > I want T.det()=1.
 >
 > Best regards,
 > Santanu


 --

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 _www: https://malb.io

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Re: [sage-devel] Unimodular transformation matrix of LLL algorithm

2020-09-27 Thread Nils Bruin 
You could do the same thing as you do with Gaussian elimination to track 
the row operations: augment the matrix with an identity matrix.
In order for the augmentation to not affect your LLL reduction, you'd want 
to multiply your original matrix by a large constant, so that the augmented 
coordinates do not affect the norms significantly.

On Sunday, September 27, 2020 at 11:20:28 AM UTC-7, Dima Pasechnik wrote:
>
>
>
> On Sun, 27 Sep 2020, 18:43 Santanu Sarkar,  > wrote:
>
>> Dear Martin,
>> Thank you so much. It works!
>> Can we make it faster? 
>> It took 17 seconds for my problem but 
>> M1.LLL() took only 3 seconds. Of course I understand we 
>> are calculating extra matrix U.  
>>
> one needs to do some Cython programming, as suggested by
> the trac ticket mentioned above.
>
>
>
>> Thanks again for your help. 
>>
>> Regards,
>> Santanu 
>>
>>
>>
>> On Sun, 27 Sep 2020 at 20:45, 'Martin R. Albrecht' via sage-devel <
>> sage-...@googlegroups.com > wrote:
>>
>>> Hi there,
>>>
>>> This should do the trick:
>>>
>>> sage: from fpylll import *
>>> sage: A = random_matrix(ZZ, 6, 90)
>>> sage: U = IntegerMatrix.identity(6)
>>> sage: B = LLL.reduction(IntegerMatrix.from_matrix(A), 
>>> U).to_matrix(matrix(ZZ, 6,
>>> 90))
>>> sage: U = U.to_matrix(matrix(ZZ, 6,6))
>>> sage: B == U*A
>>> True
>>> sage: abs(U.det())
>>> 1
>>>
>>> Cheers,
>>> Martin
>>>
>>>
>>> Santanu Sarkar > writes:
>>> > Dear all,
>>> >I have a matrix M1 with integer entries with 90 rows and 6 columns.
>>> > After applying LLL algorithm of M1, I get M2=M1.LLL(). I want to get
>>> > corresponding unimodular transformation matrix T such that
>>> > T*M1=M2. We can find T by
>>> > T=M2*M1.pseudoinverse() or T== M1.solve_left(M2), but determinant of T
>>> > becomes 0 i.e.,  T.det()=0.
>>> > I want T.det()=1.
>>> >
>>> > Best regards,
>>> > Santanu
>>>
>>>
>>> -- 
>>>
>>> _pgp: https://keybase.io/martinralbrecht
>>> _www: https://malb.io
>>>
>>> -- 
>>> You received this message because you are subscribed to the Google 
>>> Groups "sage-devel" group.
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>>> .
>>>
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>>  
>> 
>> .
>>
>

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Re: [sage-devel] Unimodular transformation matrix of LLL algorithm

2020-09-27 Thread Dima Pasechnik 
On Sun, 27 Sep 2020, 18:43 Santanu Sarkar, 
wrote:

> Dear Martin,
> Thank you so much. It works!
> Can we make it faster?
> It took 17 seconds for my problem but
> M1.LLL() took only 3 seconds. Of course I understand we
> are calculating extra matrix U.
>
one needs to do some Cython programming, as suggested by
the trac ticket mentioned above.



> Thanks again for your help.
>
> Regards,
> Santanu
>
>
>
> On Sun, 27 Sep 2020 at 20:45, 'Martin R. Albrecht' via sage-devel <
> sage-devel@googlegroups.com> wrote:
>
>> Hi there,
>>
>> This should do the trick:
>>
>> sage: from fpylll import *
>> sage: A = random_matrix(ZZ, 6, 90)
>> sage: U = IntegerMatrix.identity(6)
>> sage: B = LLL.reduction(IntegerMatrix.from_matrix(A),
>> U).to_matrix(matrix(ZZ, 6,
>> 90))
>> sage: U = U.to_matrix(matrix(ZZ, 6,6))
>> sage: B == U*A
>> True
>> sage: abs(U.det())
>> 1
>>
>> Cheers,
>> Martin
>>
>>
>> Santanu Sarkar  writes:
>> > Dear all,
>> >I have a matrix M1 with integer entries with 90 rows and 6 columns.
>> > After applying LLL algorithm of M1, I get M2=M1.LLL(). I want to get
>> > corresponding unimodular transformation matrix T such that
>> > T*M1=M2. We can find T by
>> > T=M2*M1.pseudoinverse() or T== M1.solve_left(M2), but determinant of T
>> > becomes 0 i.e.,  T.det()=0.
>> > I want T.det()=1.
>> >
>> > Best regards,
>> > Santanu
>>
>>
>> --
>>
>> _pgp: https://keybase.io/martinralbrecht
>> _www: https://malb.io
>>
>> --
>> You received this message because you are subscribed to the Google Groups
>> "sage-devel" group.
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>> email to sage-devel+unsubscr...@googlegroups.com.
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>> https://groups.google.com/d/msgid/sage-devel/87h7rjmesx.fsf%40googlemail.com
>> .
>>
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> 
> .
>

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Re: [sage-devel] Unimodular transformation matrix of LLL algorithm

2020-09-27 Thread Santanu Sarkar 
Dear Martin,
Thank you so much. It works!
Can we make it faster?
It took 17 seconds for my problem but
M1.LLL() took only 3 seconds. Of course I understand we
are calculating extra matrix U.

Thanks again for your help.

Regards,
Santanu



On Sun, 27 Sep 2020 at 20:45, 'Martin R. Albrecht' via sage-devel <
sage-devel@googlegroups.com> wrote:

> Hi there,
>
> This should do the trick:
>
> sage: from fpylll import *
> sage: A = random_matrix(ZZ, 6, 90)
> sage: U = IntegerMatrix.identity(6)
> sage: B = LLL.reduction(IntegerMatrix.from_matrix(A),
> U).to_matrix(matrix(ZZ, 6,
> 90))
> sage: U = U.to_matrix(matrix(ZZ, 6,6))
> sage: B == U*A
> True
> sage: abs(U.det())
> 1
>
> Cheers,
> Martin
>
>
> Santanu Sarkar  writes:
> > Dear all,
> >I have a matrix M1 with integer entries with 90 rows and 6 columns.
> > After applying LLL algorithm of M1, I get M2=M1.LLL(). I want to get
> > corresponding unimodular transformation matrix T such that
> > T*M1=M2. We can find T by
> > T=M2*M1.pseudoinverse() or T== M1.solve_left(M2), but determinant of T
> > becomes 0 i.e.,  T.det()=0.
> > I want T.det()=1.
> >
> > Best regards,
> > Santanu
>
>
> --
>
> _pgp: https://keybase.io/martinralbrecht
> _www: https://malb.io
>
> --
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> "sage-devel" group.
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> .
>

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Re: [sage-devel] Re: possible bug with manifolds

2020-09-27 Thread Abhishek cherath 
Funny... it works after restarting the kernel. I wonder how I managed that.

On Sun, Sep 27, 2020 at 8:22 AM Eric Gourgoulhon 
wrote:

> I cannot reproduce the issue with Sage 9.1. Are you sure that you have a
> working install of Sage?
>
> Le dimanche 27 septembre 2020 à 11:03:25 UTC+2, abhic...@gmail.com a
> écrit :
>
>>
>> I was following this
>> 
>> tutorial, and ran into issues with set_default_chart
>>
>>  minimal example:
>>
>> E. = EuclideanSpace(coordinates='spherical')
>> cartesian. = E.cartesian_coordinates()
>> E.set_default_chart(cartesian)
>>
>> error:
>>
>>
>> /home/abhishek/Applications/SageMath/local/lib/python3.7/site-packages/sage/manifolds/chart.py
>> in __call__(self, point) 531 532 """ --> 533 return point.coord(self) 534
>> 535 def domain(self): AttributeError: 'RealDiffChart' object has no
>> attribute 'coord'
>>
>> version:
>> 횂횊횐횎홼횊횝횑횟횎횛횜횒횘횗퟿.ퟷ,횁횎횕횎횊횜횎홳횊횝횎:ퟸퟶퟸퟶ⎯ퟶퟻ⎯ퟸퟶ
>> 64bit.
>>
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>

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[sage-devel] Python 3.9 support ticket needs review

2020-09-27 Thread Matthias Koeppe 
Python 3.9 is expected to be released in 1 week (2020-10-05). 
Ticket https://trac.sagemath.org/ticket/30184 adds support for it and is 
waiting for review.

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Re: [sage-devel] Unimodular transformation matrix of LLL algorithm

2020-09-27 Thread 'Martin R. Albrecht' via sage-devel 
Hi there,

This should do the trick:

sage: from fpylll import *
sage: A = random_matrix(ZZ, 6, 90)
sage: U = IntegerMatrix.identity(6)
sage: B = LLL.reduction(IntegerMatrix.from_matrix(A), U).to_matrix(matrix(ZZ, 6,
90))
sage: U = U.to_matrix(matrix(ZZ, 6,6))
sage: B == U*A
True
sage: abs(U.det())
1

Cheers,
Martin


Santanu Sarkar  writes:
> Dear all,
>I have a matrix M1 with integer entries with 90 rows and 6 columns.
> After applying LLL algorithm of M1, I get M2=M1.LLL(). I want to get
> corresponding unimodular transformation matrix T such that
> T*M1=M2. We can find T by
> T=M2*M1.pseudoinverse() or T== M1.solve_left(M2), but determinant of T
> becomes 0 i.e.,  T.det()=0.
> I want T.det()=1.
>
> Best regards,
> Santanu


-- 

_pgp: https://keybase.io/martinralbrecht
_www: https://malb.io

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[sage-devel] Re: Unimodular transformation matrix of LLL algorithm

2020-09-27 Thread Samuel Lelievre 
There is a ticket for that with some hints on how to do it.

- Sage Trac ticket 25191
  Add flag for returning LLL transformation matrix
  https://trac.sagemath.org/ticket/25191

Hopefully someone can expand on the hints, or even
push a branch to the ticket.

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[sage-devel] Unimodular transformation matrix of LLL algorithm

2020-09-27 Thread Santanu Sarkar 
Dear all,
   I have a matrix M1 with integer entries with 90 rows and 6 columns.
After applying LLL algorithm of M1, I get M2=M1.LLL(). I want to get
corresponding unimodular transformation matrix T such that
T*M1=M2. We can find T by
T=M2*M1.pseudoinverse() or T== M1.solve_left(M2), but determinant of T
becomes 0 i.e.,  T.det()=0.
I want T.det()=1.

Best regards,
Santanu

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[sage-devel] Re: possible bug with manifolds

2020-09-27 Thread Eric Gourgoulhon 
I cannot reproduce the issue with Sage 9.1. Are you sure that you have a 
working install of Sage?

Le dimanche 27 septembre 2020 à 11:03:25 UTC+2, abhic...@gmail.com a écrit :

>
> I was following this 
> 
>  
> tutorial, and ran into issues with set_default_chart
>
>  minimal example:
>
> E. = EuclideanSpace(coordinates='spherical')
> cartesian. = E.cartesian_coordinates()
> E.set_default_chart(cartesian)
>
> error:
>
>
> /home/abhishek/Applications/SageMath/local/lib/python3.7/site-packages/sage/manifolds/chart.py
>  
> in __call__(self, point) 531 532 """ --> 533 return point.coord(self) 534 
> 535 def domain(self): AttributeError: 'RealDiffChart' object has no 
> attribute 'coord'
>
> version: 
> 횂횊횐횎홼횊횝횑횟횎횛횜횒횘횗퟿.ퟷ,횁횎횕횎횊횜횎홳횊횝횎:ퟸퟶퟸퟶ⎯ퟶퟻ⎯ퟸퟶ
> 64bit.
>

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[sage-devel] possible bug with manifolds

2020-09-27 Thread Abhishek cherath 

I was following this 

 
tutorial, and ran into issues with set_default_chart

 minimal example:

E. = EuclideanSpace(coordinates='spherical')
cartesian. = E.cartesian_coordinates()
E.set_default_chart(cartesian)

error:

/home/abhishek/Applications/SageMath/local/lib/python3.7/site-packages/sage/manifolds/chart.py
 
in __call__(self, point) 531 532 """ --> 533 return point.coord(self) 534 
535 def domain(self): AttributeError: 'RealDiffChart' object has no 
attribute 'coord'

version: 
횂횊횐횎홼횊횝횑횟횎횛횜횒횘횗퟿.ퟷ,횁횎횕횎횊횜횎홳횊횝횎:ퟸퟶퟸퟶ⎯ퟶퟻ⎯ퟸퟶ
64bit.

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